Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;



int main() {
#ifdef HOME
    assert(freopen("in", "r", stdin));
#endif


    int n;
    while (scanf("%d", &n) == 1) {
        vector<int> a(n);
        vector<int> pos(n);
        for (int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            --a[i];
            pos[a[i]] = i;
        }

        vector<int> frst(1, n - 1);
        for (int i = n - 2; i >= 0; i--) {
            bool nw = 0;
            if (frst.size() == 1) {
                nw = pos[i] > pos[frst.back()];
            } else {
                int p1 = pos[i];
                int p2 = pos[frst.back()];
                int p3 = pos[frst[(int)frst.size() - 2]];

                if ((p2 > p3) != (p1 > p3)) nw = p1 < p2;
                else nw = p1 > p2;
            }
            if (nw) frst.push_back(i);
            else frst.back()--;
        }
        printf("%d\n", (int)frst.size() - 1);
    }


#ifdef HOME
    cerr << "time: " << clock() * 1.0 / CLOCKS_PER_SEC << endl;
#endif
    return 0;
}